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In a book I am reading, the authors have calculated the Pearson $\chi^2$ statistic value as 11.50 with degrees of freedom = 9.

The p-value stated is 0.241.

I looked it up in a $\chi^2$ table but I only can see 4.168 (p=0.9) and 14.68 (p=0.1).

So 11.50 lies somewhere in between (0.241). How does the author get 0.241?

I thought we have to integrate the $\chi^2$ curve to get this exact value, or either that he just took it from a computer program result?

Or is there a quick way to get an estimate of the "exact" p-value from the table?

Glen_b
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dorothy
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1 Answers1

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The actual distribution of the test statistic in the usual multinomial goodness of fit and independencence tests in contingency tables is discrete, the chi-square distribution is an asymptotic approximation.

Usually the authors will just use a chi-squared distribution p-value from a package. Most stats packages do this for you.

You can even get it from Excel.

You can closely approximate it from tables via suitable interpolation. However, the tables would need to be more extensive than the tables you appear to have.

You can also use numerical integration to obtain approximate values. In the case of a chi-square with 9 d.f. it's perhaps just feasible to do algebraic integration (e.g. using integration by parts repeatedly to get down to something that can be handled more directly).

In this case the value doesn't exactly match chi-square functions - the chi-square p-value for 11.5 on 9 df is 0.243

The author might have done a fisher exact test (whose p-values may be close), or as noted in comments, the 11.5 is likely just rounded off but the p-value computed on the unrounded value (11.53 would be about right to get p= 0.241).

More details - like a complete reference - could help pin that down.

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Glen_b
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    +1. A note on your last points -- It seems most likely that the 11.5 as reported would be rounded ... E.g. A chi-squared statistic of 11.53 would return a p-value of 0.2411. (But the OP does say 11.50...) – James Stanley Mar 08 '13 at 06:03
  • @JamesStanley Yes, indeed, that's quote possible and could explain the difference. – Glen_b Mar 08 '13 at 07:13